How do you solve Separate Squares II on LeetCode?

Separate Squares II (LeetCode #3454) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Overlapping areas should be counted only once.

What is the time complexity of Separate Squares II?

The optimal solution for Separate Squares II runs in O(n log n) time and uses O(n) space. The complexity is dominated by the sorting of events, which is O(n log n), while maintaining the active area is linear.

What is the brute force solution for Separate Squares II?

The brute force approach for Separate Squares II is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach checks every possible horizontal line to find where the area above and below is equal. It involves calculating the area covered by squares for each line, which can be inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Separate Squares II?

Separate Squares II uses the following concepts: Array, Binary Search, Segment Tree, Sweep Line. The recommended patterns to study are: Sweep Line, Segment Tree.

What are the interview tips for Separate Squares II?

Explain your thought process clearly. Consider edge cases like overlapping squares. Practice similar problems to build confidence.

What are common mistakes when solving Separate Squares II?

Not considering overlapping areas correctly. Forgetting to check all unique y-coordinates.

#3454

Separate Squares II

Hard

ArrayBinary SearchSegment TreeSweep LineSweep LineSegment Tree

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

This approach uses a sweep line technique combined with a segment tree to efficiently calculate the area covered by squares. It allows us to find the minimum y-coordinate where the areas above and below are equal without checking every possible line.

⚙️

Algorithm

3 steps

1Step 1: Create events for the start and end of each square and sort them by y-coordinate.
2Step 2: Use a segment tree to maintain the active areas as we sweep through the y-coordinates.
3Step 3: Find the y-coordinate where the area above equals the area below using the segment tree.

solution.py16 lines

1def optimal(squares):
2    events = []
3    for x, y, l in squares:
4        events.append((y, l, 1))
5        events.append((y + l, l, -1))
6    events.sort()
7    total_area = sum(l * l for _, l, _ in squares)
8    active_area = 0
9    for y, l, typ in events:
10        if typ == 1:
11            active_area += l * l
12        else:
13            active_area -= l * l
14        if active_area * 2 == total_area:
15            return y
16    return -1

ℹ

Complexity note: The complexity is dominated by the sorting of events, which is O(n log n), while maintaining the active area is linear.

1Overlapping areas should be counted only once.
2The line must lie within one of the squares.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.